(a) The matrix \(\begin{pmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{pmatrix}\) represents a rotation about the origin through angle \(2\theta\). The transformation sequence is a shear in the x-direction followed by this rotation.

(b) The inverse of \(\mathbf{M}\) is given by \(\mathbf{M}^{-1} = \begin{pmatrix} 1 & -k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{pmatrix}\).

(c) To find when \(\mathbf{M} - \mathbf{I}\) is singular, calculate the determinant: \(\det(\mathbf{M} - \mathbf{I}) = (\cos 2\theta - 1)(k \sin 2\theta + \cos 2\theta - 1) - k \sin 2\theta \cos 2\theta + \sin^2 2\theta = 0\). Simplifying gives \(2 - 2\cos 2\theta - k \sin 2\theta = 0\), leading to \(4\sin^2 \theta = 2k \sin \theta \cos \theta\). Using double angle formulas, \(\tan \theta = \frac{1}{2}k\).

(d) Given \(k = 2\sqrt{3}\) and \(\theta = \frac{1}{3}\pi\), the matrix \(\mathbf{M}\) becomes \(\begin{pmatrix} -\frac{1}{2} & -\frac{2\sqrt{3}}{3} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}\). Transforming \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} x \\ y \end{pmatrix}\) gives \(\begin{pmatrix} -\frac{1}{2}x - \frac{2\sqrt{3}}{3}y \\ \frac{\sqrt{3}}{2}x - \frac{1}{2}y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\). Solving gives \(\sqrt{3}x + 3y = 0\).

(a) The matrix \(\mathbf{M}\) can be decomposed into two transformations: a one-way stretch with scale factor 3 parallel to the x-axis, followed by a rotation anticlockwise about the origin through an angle \(\theta\).

(b) The matrix \(\mathbf{M} = \begin{pmatrix} 3\cos \theta & -\sin \theta \\ 3\sin \theta & \cos \theta \end{pmatrix}\) transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} 3x\cos \theta - y\sin \theta \\ 3x\sin \theta + y\cos \theta \end{pmatrix}\).

For an invariant line through the origin, \(y = mx\) transforms to \(Y = mX\), leading to the equation:

\(3x\sin \theta + mx\cos \theta = 3mx\cos \theta - m^2x\sin \theta\)

\(m^2 \sin \theta - 2m \cos \theta + 3 \sin \theta = 0\)

Setting the discriminant equal to zero:

\(4\cos^2 \theta - 12\sin^2 \theta = 0\)

Using \(\tan \theta = \pm \frac{1}{\sqrt{3}}\) or \(4\cos^2 \theta - 12(1 - \cos^2 \theta) = 0\) leads to \(16\cos^2 \theta - 12 = 0\) leading to \(\cos \theta = \pm \frac{\sqrt{3}}{2}\).

Thus, \(\theta = \frac{1}{6} \pi\) and \(\theta = \frac{5}{6} \pi\).

Answer:

1. (a) A one-way stretch in the \(x\)-direction with scale factor \(a\), followed by a shear parallel to the \(x\)-axis with shear factor \(b\).

2. (b) The required matrix is \(\mathbf M^{-1}=\dfrac{1}{a}\begin{pmatrix}1&-b\\0&a\end{pmatrix}\).

3. (c) \(a=2\) and \(b=3\).

1. (a) The matrix is written as \(\mathbf M=\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&1\end{pmatrix}\). For column vectors, the right-hand matrix acts first.

   The matrix \(\begin{pmatrix}a&0\\0&1\end{pmatrix}\) maps \((x,y)\) to \((ax,y)\), so it is a one-way stretch in the \(x\)-direction with scale factor \(a\).

   The matrix \(\begin{pmatrix}1&b\\0&1\end{pmatrix}\) maps \((x,y)\) to \((x+by,y)\), so it is a shear parallel to the \(x\)-axis with shear factor \(b\).

   Therefore the stretch is applied first, followed by the shear.

2. (b) First multiply the matrices:

   \(\mathbf M=\begin{pmatrix}1&b\\0&1\end{pmatrix}\begin{pmatrix}a&0\\0&1\end{pmatrix}=\begin{pmatrix}a&b\\0&1\end{pmatrix}\).

   The matrix which maps the transformed parallelogram back to the unit square is the inverse matrix \(\mathbf M^{-1}\).

   For \(\begin{pmatrix}a&b\\0&1\end{pmatrix}\), the determinant is \(a\), so

   \(\mathbf M^{-1}=\dfrac{1}{a}\begin{pmatrix}1&-b\\0&a\end{pmatrix}\).

3. (c) The area scale factor of a matrix transformation is the absolute value of its determinant. Since \(a\) and \(b\) are positive,

   \(\det\mathbf M=a\).

   The unit square has area \(1\), and its image has area \(2\), so \(a=2\).

   Now \(\mathbf M\) maps \((x,y)\) to \((X,Y)\), where

   \(\begin{pmatrix}X\\Y\end{pmatrix}=\begin{pmatrix}a&b\\0&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}ax+by\\y\end{pmatrix}\).

   The line \(x+3y=0\) has \(y=-\dfrac{x}{3}\). Substituting this into the image coordinates gives

   \(X=ax+b\left(-\dfrac{x}{3}\right)=\left(a-\dfrac{b}{3}\right)x\), and \(Y=-\dfrac{x}{3}\).

   For the line to be invariant, the image point must also lie on the same line, so \(X+3Y=0\):

   \(\left(a-\dfrac{b}{3}\right)x+3\left(-\dfrac{x}{3}\right)=0\).

   Thus \(\left(a-\dfrac{b}{3}-1\right)x=0\), so \(a-\dfrac{b}{3}=1\).

   Using \(a=2\),

   \(2-\dfrac{b}{3}=1\), so \(b=3\).

   Therefore \(a=2\) and \(b=3\).

Answer: (i) \(\alpha=15\). A basis for the range space is \(\left\{\begin{pmatrix}1\\5\\4\\-2\end{pmatrix},\begin{pmatrix}2\\2\\0\\4\end{pmatrix}\right\}\).

(ii) A basis for the null space is \(\left\{\begin{pmatrix}-2\\1\\8\\0\end{pmatrix},\begin{pmatrix}14\\-23\\0\\8\end{pmatrix}\right\}\).

Reduce the matrix \(M\):

\(\begin{pmatrix}1&2&0&4\\5&2&1&-3\\4&0&1&-7\\-2&4&-1&\alpha\end{pmatrix}\to\begin{pmatrix}1&2&0&4\\0&-8&1&-23\\0&-8&1&-23\\0&8&-1&\alpha+8\end{pmatrix}\to\begin{pmatrix}1&2&0&4\\0&-8&1&-23\\0&0&0&0\\0&0&0&\alpha-15\end{pmatrix}\).

Since the rank is \(2\), the last row must be zero. Hence

\(\alpha-15=0\), so \(\alpha=15\).

With \(\alpha=15\), two independent columns of \(M\) are the first two columns, so a basis for the range space is

\(\left\{\begin{pmatrix}1\\5\\4\\-2\end{pmatrix},\begin{pmatrix}2\\2\\0\\4\end{pmatrix}\right\}\).

For the null space, solve \(M\begin{pmatrix}x\\y\\z\\t\end{pmatrix}=0\). The reduced equations are

\(x+2y+4t=0\),

\(-8y+z-23t=0\).

Let \(t=\lambda\) and \(z=\mu\). Then

\(y=\dfrac18\mu-\dfrac{23}{8}\lambda\),

and

\(x=-2y-4t=-\dfrac14\mu+\dfrac74\lambda\).

Choosing \((\mu,\lambda)=(8,0)\) and \((0,8)\), a basis for the null space is

\(\left\{\begin{pmatrix}-2\\1\\8\\0\end{pmatrix},\begin{pmatrix}14\\-23\\0\\8\end{pmatrix}\right\}\).

Answer: (i) For \(\alpha\neq0\), \(K_1=\operatorname{span}\{(0,1,0,2)\}\).

(ii) For \(\alpha=0\), one basis for \(K_2\) is \(\{(0,1,0,2),\,(-3,0,1,-3)\}\).

(iii) Yes. Since the basis vector for \(K_1\) is one of the basis vectors for \(K_2\), \(K_1\) is a subspace of \(K_2\).

Let \(\mathbf{x}=(x,y,z,t)^T\). Row reducing the coefficient matrix gives

\(\begin{pmatrix}1&2&\alpha&-1\\2&6&-3&-3\\3&10&-6&-5\end{pmatrix}\sim\begin{pmatrix}1&2&\alpha&-1\\0&2&-3-2\alpha&-1\\0&0&\alpha&0\end{pmatrix}.\)

(i) For \(\alpha\neq0\), the last row gives \(\alpha z=0\), so \(z=0\). The second row then gives

\(2y-t=0\),

so \(t=2y\). The first row gives

\(x+2y-t=0\),

so \(x=0\). Hence

\((x,y,z,t)=y(0,1,0,2).\)

Therefore

\(\boxed{K_1=\operatorname{span}\{(0,1,0,2)\}}.\)

(ii) If \(\alpha=0\), the equations reduce to

\(x+2y-t=0,\)

\(2y-3z-t=0.\)

From the second equation, \(t=2y-3z\). Substituting into the first gives

\(x+2y-(2y-3z)=0\),

so \(x=-3z\). Hence

\((x,y,z,t)=y(0,1,0,2)+z(-3,0,1,-3).\)

Thus one basis is

\(\boxed{\{(0,1,0,2),\,(-3,0,1,-3)\}}.\)

(iii) The vector \((0,1,0,2)\), which spans \(K_1\), is also a basis vector for \(K_2\). Therefore every vector in \(K_1\) lies in \(K_2\), and so

\(\boxed{K_1\text{ is a subspace of }K_2}.\)

Answer: (a) Shear with the \(x\)-axis fixed; \((0,1)\mapsto(\frac32,1)\).

(b) Shear with the \(y\)-axis fixed; \((1,0)\mapsto(1,\frac32)\).

(c) \(\det(AB)=1\), so the area scale factor is \(1\).

(d) \(y=-2x\) and \(y=\frac12x\).

(a)

Under \(A\),

\(\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}x+\frac32y\\y\end{pmatrix}.\)

All points on the \(x\)-axis have \(y=0\), so they are unchanged. The point \((0,1)\) maps to \((\frac32,1)\). Therefore the transformation is a shear with the \(x\)-axis fixed.

(b)

Under \(B\),

\(\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}x\\\frac32x+y\end{pmatrix}.\)

All points on the \(y\)-axis have \(x=0\), so they are unchanged. The point \((1,0)\) maps to \((1,\frac32)\). Therefore the transformation is a shear with the \(y\)-axis fixed.

(c)

Compute

\(AB=\begin{pmatrix}1&\frac32\\0&1\end{pmatrix}\begin{pmatrix}1&0\\\frac32&1\end{pmatrix}=\begin{pmatrix}\frac{13}{4}&\frac32\\\frac32&1\end{pmatrix}.\)

Then

\(\det(AB)=\frac{13}{4}\cdot1-\frac32\cdot\frac32=\frac{13}{4}-\frac94=1.\)

The determinant gives the area scale factor, so the two triangles have the same area.

(d)

Using

\(AB=\begin{pmatrix}\frac{13}{4}&\frac32\\\frac32&1\end{pmatrix},\)

a point \((x,y)\) maps to

\((X,Y)=\left(\frac{13}{4}x+\frac32y,\frac32x+y\right).\)

An invariant line through the origin has the form \(y=mx\). Substitute \(y=mx\) and require \(Y=mX\):

\(\frac32x+mx=m\left(\frac{13}{4}x+\frac32mx\right).\)

For \(x\ne0\),

\(\frac32+m=\frac{13}{4}m+\frac32m^2.\)

Hence

\(m^2+\frac32m-1=0=(m+2)\left(m-\frac12\right).\)

So \(m=-2\) or \(m=\frac12\), giving \(y=-2x\) and \(y=\frac12x\).

Mark scheme